The coefficient of friction between the wedge and the table is `mu`. Friction between the block and the wedge is to be neglected.

Answer

`Q=(mMg cos(alpha-varphi))/(M cos varphi +m sin alpha sin(alpha-varphi))`.

Explanation

If the wedge moves to the left, the equations of motion assume the following form:
`Q sin alpha= ma_(x), - mg + Q cos alpha = ma_(y), " "- Q sin alpha +T = Mb_(x)`
`a_(y) = (-a_(x) +b_(x)) tan alpha, " " T=mu(Q cos alpha + Mg)`
from which
`Q=(Mmg (cos alpha +mu sin alpha))/(M+m sin alpha(sin alpha-mu cos alpha))`
`a_(x)=(Mg sin alpha (cos alpha +mu sin alpha))/(M+m sin alpha (sin alpha - mu cos alpha))`
`a_(y)= - ((M+m) g sin alpha (sin alpha - mu cos alpha))/(M+m sin alpha (sin alpha - mu cos alpha))`
`b_(x)= mu g- (mg (cos alpha +mu sin alpha) (sin alpha-mu cos alpha))/(M+m sin alpha (sin alpha - mu cos alpha))`
Since `b_(x) le 0` (explain why), it follows that
`mu g le (mg (cos alpha + mu sin alpha) (sin alpha - mu cos alpha))/(M+ m sin alpha (sin alpha - mu cos alpha))`
Hence after some simple transformations we obtain that our solution is meaningful in conditions when
`mu lt (m sin alpha cos alpha)/( M + m cos^(2)alpha)`
In this case the expression
`sin alpha - mu cos alpha gt (M sin alpha)/(M + m cos^(2) alpha)`
and consequently `a_(y) lt 0`, which agrees with the problems.s idea.
If it turns out that `mu=(m sin alpha cos alpha)/(M=m cos^(2)alpha)`, then `b_(x)=0`, i.e. the wedge will move on teh table at the initial speed `v_(0)` (or remain stationary, if `v_(0) = 0`). In this case
`sin alpha - mu cos alpha= (M sin alpha)/(M+ m cos^(2)alpha), cos alpha + mu sin alpha =((M+m)cos alpha)/(M+m cos^(2)alpha)`
from which
`Q=mg cos alpha`
`a_(x)= g sin alpha cos alpha, a_(y)= - g sin^(2) alpha, a= sqrt( a_(x)^(2)+a_(y)^(2)) = g sin alpha`